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Rank and submatrices theorem

  1. Nov 11, 2011 #1
    Let A be a nonzero matrix of size n. Let a k*k submatrix of A be defined as a matrix obtained by deleting any n-k rows and n-k columns of A. Let m denote the largest integer such that some m*m submatrix has a nonzero determinant. Then rank(A) = k.

    Conversely suppose that rank(A) = m. There exists a m*m submatrix has a nonzero determinant.



    I'm currently trying to prove this theorem. Not quite sure if I should proceed by examining the solution space of A or rather just do something clever with the determinants. I feel like there's a property of determinants that I'm missing that'd make this much easier.
     
  2. jcsd
  3. Nov 11, 2011 #2
    If the determinant is non-zero, that implies that the columns are linearly independent. Remember, the determinant measures the (hyper-)volume of the parallelepiped spanned by the columns. Intuitively, a non-zero volume implies linear independence. I would use that.

    So, you have a square submatrix whose columns are linearly independent. What does that tell you about the corresponding columns in the bigger matrix A?
     
  4. Nov 11, 2011 #3

    Deveno

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    some things to think about:

    if you use row-reduction to find rank(A), how does each row-operation affect the determinant?

    if you think of the determinant as a function of n n-vectors, instead of an nxn matrix, is it linear in each variable? how does this tie in to dim(row(A)) = dim(col(A))?

    does re-arranging rows or columns of a matrix change its determinant?
     
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